Problem: We are given that $\dfrac{dy}{dx}=\sqrt{1-y^2}$. Find an expression for $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$. $\dfrac{d^2y}{dx^2}=$
Notice that the equation defines $y$ implicitly—we don't have an explicit expression for $y$ in terms of $x$. So we will have to use implicit differentiation. If we differentiate the equation once, we will have $\dfrac{d^2y}{dx^2}$ on the left-hand side of the equation. This is what we get after we differentiate each side of the equation: $\dfrac{d^2y}{dx^2}=-\dfrac{y\cdot\dfrac{dy}{dx}}{\sqrt{1-y^2}}$ The expression we got for $\dfrac{d^2y}{dx^2}$ isn't in terms of only $x$ and $y$ because it includes $\dfrac{dy}{dx}$ in it. Let's plug the original equation ${\dfrac{dy}{dx}=\sqrt{1-y^2}}$ in the new equation to obtain an expression in terms of $y$ alone: $\begin{aligned} \dfrac{d^2y}{dx^2}&=-\dfrac{y\cdot{\dfrac{dy}{dx}}}{\sqrt{1-y^2}} \\\\ &=-\dfrac{y\cdot{\sqrt{1-y^2}}}{\sqrt{1-y^2}} \\\\ &=-y \end{aligned}$ In conclusion, this is an expression for $\dfrac{d^2y}{dx^2}$ in terms of $x$ and $y$ : $\dfrac{d^2y}{dx^2}=-y$